HW1_Solution

HW1_Solution - Spring 10 ECE 35 HW1 Solutions P1.7-2 P2.2-3...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
1 Spring 10 ECE 35 HW1 Solutions P1.7-2 P2.2-3 (a) Plot the data. The data points do indeed lie on a straight line. The slope of the line is 256.5 V/A and the line passes through the origin so the equation of the line is 256.5 vi = . The element is indeed linear. Another method is to write down the linear relation as v = ki , then calculate k for each data points. They all lead to k = 256.5 V/A so it verifies the data with linear model. (b) When i = 4 mA, v = (256.5 V/A) × (4 mA) = (256.5 V/A) × (0.004 A) = 1.026 V (c) When v = 12 V, 12 0.04678 256.5 i == A = 46.78 mA. P2.4-10 Label the current i as shown. That current is the element current in both resistors. First a 40 v i = Next ab b a 40 40 vv vR i R R v = For example, 7.05 40 24 11.75 R Ω
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2 P2.5-1 (a) () 2 2 15 3 A and 5 3 45 W 5 s v iP R i R === = = = (b) and do not depend on . s i The values of and are 3 A and 45 W, both when 3 A and when 5 A. ss i i = = P2.5-2 (a) 22 10 5 2 10 V and 20 W 5 s v vR i P R == = = = = (b) and do not depend on . s vP v The values of and are 10V and 20 W both when 10 V and when 5 V v v = = P2.6-2 The voltmeter current is zero so the ammeter current is equal to the
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 09/10/2010 for the course ECE 107 taught by Professor Fullterton during the Spring '07 term at UCSD.

Page1 / 3

HW1_Solution - Spring 10 ECE 35 HW1 Solutions P1.7-2 P2.2-3...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online